Cyclic Algebra. Suppose that $a,b,c$ are real numbers satisfying $a^2 + b^2 + c^2 = 1$ and $a^3 + b^3 + c^3 = 1$.
Find all possible value(s) of $a+b+c$.
My solution :
$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow   a+b+c=1+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow a+b+c=1+c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)$
$\Rightarrow   a+b+c=1+c(1-c^2)+a(1-a^2)+b(1-b^2)$
$\Rightarrow   a+b+c=1+c-c^3+a-a^3+b-b^3$
$\Rightarrow   a+b+c=1+c+a+b-1$
$\Rightarrow   a+b+c=c+a+b$
This led me to the same expression $a+b+c$.
Is there a smarter way to solve this problem?
 A: The numbers $a,b,c$ are the roots of the cubic
$$
(x-a)(x-b)(x-c) = x^3 - e_1 x^2 + e_2 x - e_3
$$
where $e_1=a+b+c$, $e_2=ab+ac+bc$, $e_3=abc$.
Newton's identities tells us that
$$
e_{1}=p_{1},
\quad
2e_{2}=e_{1}p_{1}-p_{2},
\quad
3e_{3}=e_{2}p_{1}-e_{1}p_{2}+p_{3}
$$
for $p_1=a+b+c$, $p_2=a^2 + b^2 + c^2$, $p_3=a^3 + b^3 + c^3$.
Newton's identities with $p_1=s$, $p_2=1$, $p_3=1$ then give
$$
e_2 = \frac12(s^2-1),
\quad
e_3 = \frac16 (s - 1)^2 (s + 2)
$$
The discrimant of the cubic is then
$$
\Delta = \frac16(-s^6 + 9 s^4 - 8 s^3 - 21 s^2 + 36 s - 15)
= -\frac16 (s - 1)^2 (s^4 + 2 s^3 - 6 s^2 - 6 s + 15)
$$
The three roots of the cubic are real iff $\Delta \ge 0$ and this happens iff $s=1$ because $s^4 + 2 s^3 - 6 s^2 - 6 s + 15 \ge 0$ for all $s$.
(done with lots of help from WA)
A: For a collection $a_1,...,a_n$ of real numbers, let $m=\max|a_i|$. We notice that $m \leq \big(\sum_i |a_i|^2\big)^{1/2}$ with equality iff there exists an index $i$ such that $|a_i| = m$ and $a_j=0$ for $j \neq i$. Moreover, since $|a_i|^3 \leq m|a_i|^2$ we notice that $$\bigg( \sum_i |a_i|^3 \bigg)^{1/3} \leq \bigg( \sum_i |a_i|^2 \bigg)^{1/3} \cdot m^{1/3} \leq \bigg( \sum_i |a_i|^2 \bigg)^{1/3} \cdot \bigg(\sum_i |a_i|^2 \bigg)^{ \frac12 \cdot \frac13} = \bigg( \sum_i |a_i|^2 \bigg)^{1/2}.$$ Now if we know that $\sum a_i^2 = \sum a_i^3 =1$, then it holds that $$1 = \bigg(\sum a_i^3 \bigg)^{1/3} \leq \bigg( \sum |a_i|^3 \bigg)^{1/3} \leq \bigg( \sum a_i^2 \bigg)^{1/2} = 1,$$ and therefore equality must hold throughout both of the previous expressions. This forces one to have $m= \big(\sum a_i^2 \big)^{1/2}$. Hence some $|a_i|=m$ and all other $a_j$ are zero. This forces $a_i=1$, hence the sum must be $1$.
A: Hint: consider $$(a+b+c)^2$$ and $$(a+b+c)^3$$
we have
$$\frac{a+b+c}{3}\le \sqrt{\frac{a^2+b^2+c^2}{3}}$$
